- #1

Tio Barnabe

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- #1

Tio Barnabe

- #2

Mark44

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First off, ##\frac{d}{d \cos \theta} \sin \theta## is pretty unwieldy, as ##\sin(\theta)## isn't a function of ##\cos(\theta)##

Second, integration and differentiation are different operations, so the fact that the integral of some function on some interval is divergent doesn't have any bearing here.

- #3

FactChecker

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- #4

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Using the transformation ##\xi(\theta)=\text{cos}(\theta)##

$$\text{sin}(\theta)=\pm\sqrt{1-\xi^{2}}$$

so

$$\frac{d}{d\text{cos}(\theta)}\text{sin}(\theta)=\pm\frac{d}{d\xi}\sqrt{1-\xi^{2}}=\pm\frac{\xi}{\sqrt{1-\xi^{2}}}=\pm\frac{\text{cos}(\theta)}{\text{sin}(\theta)}=\pm\text{cot}(\theta)$$

- #5

Tio Barnabe

Thank you to everyone

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